= Solution
For $\varepsilon\ne0$, use $z=e^{i\theta}$. The interval traverses $|z|=1$ once counterclockwise, $\cos\theta=(z+z^{-1})/2$, and $d\theta=dz/(iz)$. Thus
$$
I(\varepsilon)=\oint_{|z|=1}\frac{2\,dz}{i(\varepsilon z^2+2z+\varepsilon)}.
$$
Put $d=\sqrt{1-\varepsilon^2}>0$. The two simple poles are
$$
z_+=\frac{-1+d}{\varepsilon}=-\frac{\varepsilon}{1+d},\qquad z_-=-\frac{1+d}{\varepsilon}.
$$
Their moduli are $|\varepsilon|/(1+d)<1$ and $(1+d)/|\varepsilon|>1$, so exactly $z_+$ is enclosed, for either sign of $\varepsilon$. Its <residue> is
$$
\frac{2}{i(2\varepsilon z_++2)}=\frac1{id}.
$$
The <residue theorem> yields
$$
\boxed{I(\varepsilon)=\frac{2\pi}{\sqrt{1-\varepsilon^2}},\qquad -1<\varepsilon<1.}
$$
This also follows from the <reciprocal affine sine integral> after an angular shift. At $\varepsilon=0$, the original integral is directly $2\pi$, agreeing with the boxed formula. \b[The displayed two-pole formulas require $\varepsilon\ne0$]: at zero the contour integrand reduces to $1/(iz)$ with a single pole. Thus the source's claim of two simple poles has this removable special-case exception.
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